ServletForwardingController

Spring Controller implementation that forwards to a named servlet, i.e. the "servlet-name" in web.xml rather than a URL path mapping. A target servlet doesn't even need a "servlet-mapping" in web.xml in the first place: A "servlet" declaration is sufficient.

Useful to invoke an existing servlet via Spring's dispatching infrastructure, for example to apply Spring HandlerInterceptors to its requests. This will work even in a minimal Servlet container that does not support Servlet filters.

Example: web.xml, mapping all "/myservlet" requests to a Spring dispatcher. Also defines a custom "myServlet", but without servlet mapping.

 <servlet> <servlet-name>myServlet</servlet-name> <servlet-class>mypackage.TestServlet</servlet-class> </servlet> <servlet> <servlet-name>myDispatcher</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> </servlet> <servlet-mapping> <servlet-name>myDispatcher</servlet-name> <url-pattern>/myservlet</url-pattern> </servlet-mapping>

 <bean id="urlMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping"> <property name="interceptors"> <list> <ref bean="openSessionInViewInterceptor"/> </list> </property> <property name="mappings"> <props> <prop key="/myservlet">myServletForwardingController</prop> </props> </property> </bean> <bean id="myServletForwardingController" class="org.springframework.web.servlet.mvc.ServletForwardingController"> <property name="servletName"><value>myServlet</value></property> </bean>

Author
Juergen Hoeller

Since
1.1.1

See Also
ServletWrappingControllerorg.springframework.orm.jpa.support.OpenEntityManagerInViewInterceptororg.springframework.orm.jpa.support.OpenEntityManagerInViewFilter

Constructors

Functions